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Find a limit of this expression $$\frac{(1+x)^\frac14-1}{(1+x)^\frac13-1}$$ as $x$ tends to $0$.

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    $\frac{3}{4}$ since $$\lim_{x\to 0}\frac{(1+x)^{\alpha}-1}{x}=\alpha $$ by Bernoulli's inequality, De l'Hopital, a change of variables or whatever. – 2017-01-29
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    See [binomial series](http://en.wikipedia.org/wiki/Binomial_series). – 2017-01-29

3 Answers 3

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Hint:

If $u = (x+1)^{\frac{1}{12}}$, we have $$\lim_{x\rightarrow 0}\frac{(1+x)^\frac14-1}{(1+x)^\frac13-1} = \lim_{u\rightarrow 1}\frac{u^3-1}{u^4-1}$$

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By applying De L'Hopital rule $$\lim _{x\to \:0}\left(\frac{\left(1+x\right)^{\frac{1}{4}}-1}{\sqrt[3]{1+x}-1}\right) =_H \lim _{x\to \:0}\left(\frac{\frac{1}{4\left(x+1\right)^{\frac{3}{4}}}}{\frac{1}{3\left(x+1\right)^{\frac{2}{3}}}}\right)= 3/4$$ Or by asymptotic approximations $$\lim _{x\to \:0}\left(\frac{\left(1+x\right)^{\frac{1}{4}}-1}{\sqrt[3]{1+x}-1}\right) \approx_0 \lim _{x\to 0}\left(\frac{\left(1+\frac{1}{4}x\:-1\right)}{\left(1+\frac{1}{3}x\:-1\right)}\right) =\frac{3}{4}$$

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$$\lim{x \to 0} \frac{(1+x)^\frac14-1}{(1+x)^\frac13-1}$$

$$ = \lim{x \to 0} \frac{(1+x)^\frac14-1}{x} \cdot \frac{x}{(1+x)^\frac13-1}$$

But $$\lim{x \to 0} \frac{(1+x)^n − 1}{x} = n$$

$$ = \frac 14 × \frac 31 = \frac 34$$